Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises 16.3 - Page 967: 21



Work Step by Step

Since, $\nabla f =\dfrac{\partial f}{ \partial x} i+\dfrac{\partial f}{ \partial y}j +\dfrac{\partial f}{ \partial z} k ...(1)$ We have: $ \dfrac{\partial f}{ \partial x}= \dfrac{1}{y} ; \dfrac{\partial f}{ \partial y}=\dfrac{-x}{y^2} +\dfrac{1}{z} \\ \dfrac{\partial f}{ \partial z}=\dfrac{-y}{z^2}$ Now, $f=\dfrac{x}{y}+g(y,z) ...(2)$ $\implies \dfrac{\partial g(y,z)}{ \partial x}=0 $ Equation (2) becomes: $f=\dfrac{x}{y}+\dfrac{y}{z}+h(z) .....(3)$ and $h(z)=C=0$ Substitute all the above values in the given integral as follows: $ \int_{1,1,1}^{2,2,2} \dfrac{1}{y} dx +(\dfrac{1}{z} -\dfrac{x}{y^2}) dy -\dfrac{y}{z^2} dz=f(2,1,1)-f(1,2,1)=(\dfrac{2}{2}+\dfrac{2}{2}+C)-(\dfrac{1}{1}+\dfrac{1}{1}+C) \\ =0$
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